CURVES IN GRASSMANNIANS

CURVES IN GRASSMANNIANS by Montserrat Teixidor i Bigas INTRODUCTION Given a curve C and a line bundle L with sections, one can consider the rational map

arXiv:alg-geom/9705020v1 22 May 1997

C → P(H 0 (L)) from the curve to projective space associated to the complete linear series of L. A great deal is known about the image curve when the degree of the line bundle is high enough. For example, the curve is projectively normal if degL ≥ 2g + 1 (cf.[C],[M]). Saint Donat(cf[SD]) proved that the ideal of the curve is generated by quadrics and cubics and if degL ≥ 2g + 2, it is generated by quadrics alone. In the same vein, one can define condition Np for a curve. Roughly, it means that the curve is projectively normal and in a minimal resolution of the ideal generators have the smallest possible degree up to the p−th term. Green (cf[G]) showed that if degL ≥ 2g + p + 1, then the curve satisfies property Np . Consider now a vector bundle E of degree d and rank n ≥ 2 on the curve. Assume that it has enough sections. We obtain a map C → G(n, H 0 (E)) where G denotes the Grassmannian. In contrast with the line bundle case, nothing seeems to be known about the geometry of the image curve in such a Grassmanian, (say under good conditions for E). In this paper, we try to clarify the situation somehow. Our main results are in Propositions (1.1),(1.2). We consider the composition f of the following two maps C → G(n, H 0 (E)) → P(∧n H 0 (E)) where the last map is the Plucker immersion of the Grassmannian in projective space. In (1,1), we show that for very small degree the image of f is a non-degenerate curve in projective space. In (1.2) we show that for sufficiently high degree the image of f lies on a proper linear subvariety and satisfies property Np as a curve in this variety. The author is a member of the Europroj group “Vector Bundles on Algebraic Curves”. The author wants to thank the Mathematics Department of Cambridge University where this paper was written. SET UP OF THE PROBLEM AND PROOF OF THE FIRST RESULTS We want to look at a curve from an external point of view. We need first decide on the set up. Most interesting vector bundles are stable or at least semistable. In contrast with line bundles, even stable vector bundles of high degree don’t behave all in the same way. We shall restrict our attention to properties that hold for the generic vector bundle. As we want to define a map to the Grassmanian, we need h0 (E) ≥ n + 2. If the vector bundle is generic, this implies d ≥ ng + 1. The map f is given by means of a (not necessarily complete) linear series. The line bundle that gives rise to this linear series is the determinant of the vector bundle E. The subspace of the space of sections is the image of the map ψ : ∧n H 0 (E) → H 0 (∧n E). The first question that we want to ask is whether the curve is degenerate in projective space (i.e. lies on some hyperplane). The answer is yes for sufficiently high degree. In fact the curve is degenerate in projective space when the map ψ above is not injective. For a generic vector bundle of degree d ≥ n(g − 1), h1 (E) = 0.
Therefore, h0 (E) = d + n(1 − g) and dim ∧n H 0 (E) = d+n(1−g) . On the other hand ∧n E is a line bundle n 0 n n 0 and h (∧ E) = d+ 1 − g. Hence, when d is large, dim∧ √H (E) ≥ h0 (∧n E) and the map cannot be injective. For example, in rank two, injectivity implies d ≤ 4g−1+2 8g+1 . We do not know whether injectivity holds in the whole allowable range (for generic vector bundle). We prove in the next two paragraphs that for very small degree, the map is indeed injective. On the other hand, one can show that the map is surjective for generic E if the degree is high enough. The next two propositions 1

show our present state of knowledge. One could ask in general whether ψ always has maximal rank for a generic E. (1,1) Proposition. Let E be a generic stable vector bundle of degree ng+1 or ng+2. Then, the map ψ is injective and therefore the image by f does not lie on any proper linear subvariety. (1,2)Proposition. Let E be a generic stable vector bundle of degree d ≥ 2ng + 1. Then, the map ψ is surjective. Hence the set of hyperplanes in the ambient space of the Grassmanian cut a complete linear system on the curve. If moreover d ≥ 2g + p + 1, then the image curve satisfies property Np in the sublinear variety that it spans. Proposition (1,1) will be proved in the next two paragraphs. In order to prove (1.2), it is enough to exhibit a vector bundle of degree d for which the map is surjective. We shall do this by induction on n. For rank one, there is nothing to prove. Assume the result true for vector bundles of rank n − 1. Consider a generic extension 0→L→E→F →0 where degL = 2g. Then F is a generic vector bundle of rank n−1 and degree degF = degE−2g ≥ 2(n−1)g+1. Hence, the induction assumption applies to F and the map ∧n−1 H 0 (F ) → H 0 (∧n−1 F ) is surjective. We need to use the following Theorem of Mumford (cf.[M]). (1,3) Proposition (Mumford [M]). If L1 , L2 are line bundles degL1 ≥ 2g, degL2 ≥ 2g + 1, then the map H 0 (L1 ) ⊗ H 0 (L2 ) → H 0 (L1 ⊗ L2 ) is surjective By (1,3), the map H 0 (L) ⊗ H 0 (∧n−1 F ) → H 0 (L ⊗ ∧n−1 F ) is surjective. Moreover, as degL = 2g, h1 (L) = 0 and H 0 (E) = H 0 (L) ⊕ H 0 (F ). Also ∧n E = L ⊗ ∧n−1 F . The surjectivity of ψ then follows. The second statement follows directly from Green and Lazarsfeld‘s Theorem. (1,4) Remark In characteristic zero, the hypothesis of genericity in (1,2) is unnecessary: It was proved by Butler (cf.[B]) that if E, F are semistable vector bundles satisfying degE = d > 2g, µ(F ) ≥ 2g, then the map H 0 (E) ⊗ H 0 (F ) → H 0 (E ⊗ F ) is surjective. Hence, by induction on n, the map ⊗n H 0 (E) → H 0 (∧n E) is surjective. Moreover the tensor product of semistable vector bundles is semistable. As ⊗n E and ∧n E have the same slope, the symmetric product S n (E) is also semistable of the same slope. Hence, h1 (S n (E)) = 0 and the map H 0 (⊗n (E)) → H 0 (∧n (E)) is surjective N+1 SECTIONS In this paragraph, we prove (1.1) for d = ng + 1. Our method of proof is to degenerate the generic E to a direct sum of a generic line bundle L of degree g and a generic vector bundle F of rank n − 1 and degree (n − 1)g + 1. The map is not injective in this case but the elements in the kernel cannot be deformed to a generic infinitessimal deformation of E. The techniques used are inspired of [W] Prop.(1.2). With our assumptions, H 0 (E) = H 0 (L) ⊕ H 0 (F ) and h0 (L) = 1, ∧n H 0 (E) = (H 0 (L) ⊗ ∧n−1 H 0 (F )) ⊕ n 0 ∧ H (F ). As F is generic, it has n sections and by induction on n, we can assume that the map ∧n−1 H 0 (F ) → H 0 (∧n−1 F ). is injective. Then, as h0 (L) = 1, the map 2

H 0 (L) ⊗ (∧n−1 H 0 (F )) → H 0 (L ⊗ ∧n−1 F ) = H 0 (E) is also injective. On the other hand, as F has rank n−1 , ∧n H 0 (F ) is contained in the kernel of ψ. Therefore, the kernel of ψ in this case is ∧n H 0 (F ). Denote by sk , k = 1...n a basis for H 0 (F ). We want to see that s1 ∧...∧sn does not extend to an element in the kernel of ψǫ for a generic infinitessimal deformation Eǫ of E. Recall that an infinitessimal deformation Eǫ of E is given by an element e ∈ H 1 (E ∗ ⊗ E). Up to the choice of a suitable covering C = ∪Ui , we can represent e by a cocicle ϕij ∈ H 0 (Ui ∩ Uj , E ∗ ⊗ E). Then, Eǫ can be represented locally as E|Ui ⊕ ǫE |Ui with gluings on Ui ∩ Uj given by the matrix   Id o ϕij Id In the case E = L ⊕ F , ϕij =



fij11 fij12

fij21 fij22



where fijkl ∈ H 0 (Ui ∩ Uj , Hom(Lk , Ll )), Lt = L(resp.F ) if t = 1(resp.2). Denote by skǫ = (0, sk ) + ǫ(sk1i , sk2i ) an extension to Eǫ of the section sk . Then as ǫ2 = 0 and ∧n H 0 (F ) ⊂ Kerψ, ψǫ (s1ǫ ∧ ... ∧ snǫ ) = ǫ

n X

(−1)k sk1i ∧ (∧l6=k sl )

k=1

As F is a vector bundle of rank n-1, it is locally a direct sum of n copies of the trivial bundle. When we take such a local representation, the condition above can be written as   1 s1i s1 det  ... ...  = 0 sn1i sn where each si stands for a row of lenght n-1. This means that the first column in the matrix is a linear combination of the remaining n-1. Equivalently, there exists a local section on Ui φi : F → L such that sk1i = φi (sk ). The gluing conditions can be read as sk2j = sk2i +fij22 sk , sk1j = sk1i +fij12 sk . Replacing sk1i = φi (sk ) gives φi + fij21 = φj . We now consider the set {(ϕij , φi , s12i ...sn2i ) ∈ C 1 (E ∗ ⊗ E) ⊕ C 0 (F ∗ ⊗ L) ⊕ (C 0 (F ))n , φj = φi + fij21 , sk2j = sk2i + fij22 sk }. We can realize this set as the first hypercohomology group H of the double complex of sheaves below: C 0 (E ∗ ⊗ E) ↓ C 0 (L ⊗ F ∗ ) ⊕ (C 0 (F ))n

C 1 (E ∗ ⊗ E) ↓ → C 1 (L ⊗ F ∗ ) ⊕ C 1 (F ))n →

C 2 (E ∗ ⊗ E) ↓ → C 2 (L ⊗ F ∗ ) ⊕ (C 2 (F ))n →

The vertical maps in this diagram are induced by (π 21 , π 22 (sk )). We then have an exact sequence 0 → H 0 (E ∗ ⊗ E) → H 0 (L ⊗ F ∗ ) ⊕ (H 0 (F ))n → H → H 1 (E ∗ ⊗ E) → H 1 (L ⊗ F ∗ ) ⊕ (H 1 (F ))n Note how this last map is defined by α : H 1 (E ∗ ⊗ E) → H 1 (L ⊗ F ∗ ) ⊕ (H 1 (F ))n (ϕij ) → (fij21 , fij22 (sk )) 3

As degF = (n − 1)g + 1, h1 (F ) = 0. As degL = g, h1 (L ⊗ F ∗ ) = (n − 1)(g − 1) + 1 6= 0, Kerα = {ϕij |fij21 = 0}. Therefore, if the deformation of E does not preserve F as a subundle, s1ǫ ∧ ... ∧ snǫ ∈ / Kerψǫ This completes the proof of the statement.

N+2 SECTIONS We now prove the analogous result for a generic vector bundle of rank n with n+2 sections. We first degenerate the vector bundle E to a direct sum E = L11 ⊕F2 ⊕L41 ⊕...⊕Ln1 where subindices denote ranks degL11 = g + 1, degF2 = 2g + 1, degL41 = ... = degLn1 = g. We next degenerate F2 = L21 ⊕ L31 , degL21 = g + 1. We then obtain E = L11 ⊕ L21 ⊕ ... ⊕ Ln1 , degL11 = degL21 = g + 1, degL31 = ... = degLn1 = g We assume all vector bundles appearing in the decompositions above to be generic. Hence, h0 (L11 ) = h0 (L21 ) = 2, h0 (L31 ) = ... = h0 (Ln1 ) = 1. Then ∧n H 0 (E) = ∧2 H 0 (L11 ) ⊗ ∧2 H 0 (L21 ) ⊗ (⊕3≤i5